#$&*
course Phy 121
A 5 kg cart rests on an incline which makes an angle of 30 degrees with the horizontal. Sketch this situation with the incline rising as you move to the right and the cart on the incline. Include an x-y coordinate system with the origin centered on the cart, with the x axis directed up and to the right in the direction parallel to the incline.
The gravitational force on the cart acts vertically downward, and therefore has nonzero components parallel and perpendicular to the incline.
Sketch the x and y components of the force, as estimate the magnitude of each component.
What angle does the gravitational force make with the positive x axis, as measured counterclockwise from the positive x axis? Which is greater in magnitude, the x or the y component of the gravitational force?
answer/question/discussion:
the y component makes an angle of 60 degrees with the x axis and the x component makes an angle of 180 degrees with the x axis
#$&*
Using the definitions of the sine and cosine, find the components of the cart's weight parallel and perpendicular to the incline.
answer/question/discussion:
F_parallel= 5kg*9.8m/s^2*sin(30)= 42.44N
F_perpendicular=5kg*9.8m/s^2*cos(30)=49N
#$&*
How much elastic or compressive force must the incline exert to support the cart, and what is the direction of this force?
answer/question/discussion:
The incline must resit the force that is perpendicular to the incline so it must support 49N
#$&*
If no other force is exerted parallel to the incline, what will be the cart's acceleration?
answer/question/discussion:
F=M*a
42.44N=5kg*a
a=8.44m/s^2
#$&*
"
@&
The acceleration of a freely falling mass will have magnitude 9.8 m/s^2.
The acceleration of a cart on a 30 degree incline won't be this close to the acceleration of a freely falling object.
If sketched as indicated, the x axis will be parallel to the incline. If you start with the x axis horizontal and the y axis vertical, this will require that the x axis be rotated 30 degrees from its original direction, and that the rotation be counterclockwise.
The entire set of coordinate axes would have to be rotated, so that the y axis will also rotate 30 degrees counterclockwise.
The weight vector will stay where it originally was, so that it is no longer directed along the negative y axis.
The question now becomes, what is the angle of the weight vector, as measured in the counterclockwise direction from the x axis.
The negative y axis lies 270 degrees in the counterclockwise direction from the positive x axis.
At what angle relative to the positive x axis does the weight vector lie?
If that angle is theta, then the x and y components of the weight vector are
W_x = W cos(theta)
and
W_y = W sin(theta).
By getting the angle theta right, W_x and W_y will both have the appropriate signs, telling you not only the magnitudes of the two forces but whether each is in the positive or negative direction relative to the appropriate axis.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
*@